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arccossint怎么算
求∫x
arccos
cd(arccosx)
答:
令arccosx=t,那么x=cost 所以得到原积分=∫ cost *t dt =∫ t d(
sint
)= t *sint -∫sint dt = t *sint +cost +C =arccosx *√(1-x^2) +x +C,C为常数
csc[
arccos
(476/485)],要用几分之几的分数值表示?
答:
t =
arccos
(476/485), cost = 476/485,
sint
= 93/485, csct = 485/93
csc[
arccos
(1300/1301)],要用几分之几的分数值表示?
答:
t =
arccos
(1300/1301), cost = 1300/1301,
sint
= 51/1301, csct = 1301/51
高数积分
答:
∫x^3
arccos
xdx/(1-x^2)^1/2 =∫(cost)^3*t*1/(
sint
)*(-sint)dt (令x=cost)=-∫t*(cost)^3dt=-∫t*(cost)^2 d(sint)=-∫t[1-(sint)^2]d(sint)=-∫td(sint)+∫t*(sint)^2d(sint)=-t*sint+∫sintdt+1/3*∫t d[(sint)^3]=-tsint-cost+1/3[t(sint)^3...
x=t-
sint
y=1-cost
如何
建立y=f(x)的方程呢?
答:
x=t-
sint
y=1-cost t=
arccos
(1-y)x=arccos(1-y)-sin[arccos(1-y)]=arccos(1-y)-√(2y-y²)x+√(2y-y²)-arccos(1-y)=0 (本参数方程只能化成y=f(x)的反函数形式)
x= r(t-
sint
), y=?
答:
x=r(t-
sint
)...(1)y=r(1-cost)...(2)由(2)得cost=1-(y/r),∴t=
arccos
[1-(y/r)]...(3);sint=sin[arccos(1-y/r)]=√[1-(1-y/r)²]=√(2y/r-y²/r²)=(1/r)√(2ry-y²)...(4)将(3)(4)代入(1)时即得:x=rarccos[1-(y/r)]...
用分部积分法求下列不定积分 1)∫xsin2xdx 2)∫xlnxdx 3)∫
arccos
...
答:
2)3)4)答案同楼上,1)∫xsin2xdx=(-1/2)∫xdcos2x=(-1/2)xcos2x+(1/2)∫cos2xdx=(-1/2)xcos2x+(1/4)sin2x+C 2)∫xlnxdx=(1/2)∫lnxdx^2=(1/2)x^2lnx-(1/2)∫xdx=(1/2)x^2lnx-(1/4)x^2+C 3)∫
arccos
xdx=xarccosx-∫-xdx/√(1-x^2)=xarctanx-...
如何
用代数的方法求函数的极小值?
答:
这个要转化 ∫√(4-x^2)dx 设x=2cost a -->cost=x/2 -->t=
arccos
(x/2)4-x^2=4sin^2t √(4-x^2)=2
sint
b -->sint=√(4-x^2) /2 dx=-2sintdt c 则原代化为 ∫2sint*(-2sintdt)=2∫-2sin^2tdt =2∫(cos2t-1)dt =2∫cos2t dt -2∫dt =...
定积分根号下2-x^2的原函数
怎么
求? 上限是根号2,下限是1
答:
∫(1->√2)√(2-x^2)dx 令x=√2cost √(2-x^2)=√2
sint
dx=-√2sintdt 先不忙管上下限 ∫√2sint*-√2sintdt =∫(-2sin^2t) dt =∫(cos2t-1)dt =∫cos2tdt-∫dt =sin2t /2 -t =sintcost-t =1/2*√2sint*√2cost -t =1/2*√(2-x^2)*x-
arccos
(x/√2)...
∫√(4-x^2)dx等于?
答:
这个要转化 ∫√(4-x^2)dx 设x=2cost a -->cost=x/2 -->t=
arccos
(x/2)4-x^2=4sin^2t √(4-x^2)=2
sint
b -->sint=√(4-x^2) /2 dx=-2sintdt c 则原代化为 ∫2sint*(-2sintdt)=2∫-2sin^2tdt =2∫(cos2t-1)dt =2∫cos2t dt -2∫dt =...
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